T T T I I I M M M E E E S S S A A A V V V I I I N N N G G G D D D E E E S S S I I I G G G N N N A A A I I I D D D S S S Columns Portland Cement Association Page 1 of 9 The following examples illustrate the design methods presented in the article “Timesaving Design Aids for Reinforced Concrete, Part 3: Columns and Walls,” by David A. Fanella, which appeared in the November 2001 edition of Structural Engineer magazine. Unless otherwise noted, all referenced table, figure, and equation numbers are from that article. The examples presented here are for columns. Examples for walls are available on our Web page: www.portcement.org/buildings. Example 1 In this example, an interior column at the 1 st floor level of a 7-story building is designed for the effects of gravity loads. Structural walls resist lateral loads, and the frame is nonsway. Materials • Concrete: normal weight (150 pcf), ¾-in. maximum aggregate, f′ c = 5,000 psi • Mild reinforcing steel: Grade 60 (f y = 60,000 psi) Loads • Floor framing dead load = 80 psf • Superimposed dead loads = 30 psf • Live load = 100 psf (floor), 20 psf (roof) Building Data • Typical interior bay = 30 ft x 30 ft • Story height = 12 ft-0 in. The table below contains a summary of the axial loads due to gravity. The total factored load P u is computed in accordance with Sect. 9.2.1, and includes an estimate for the weight of the column. Live load reduction is determined from ASCE 7-98. Moments due to gravity loads are negligible. Floor DL (psf) LL (psf) Red. LL (psf) P u (kips) Cum. P u (kips) 7 80 20 20.0 142 142 6 120 100 50.0 238 380 5 120 100 42.7 227 607 4 120 100 40.0 223 830 3 120 100 40.0 223 1,053 2 120 100 40.0 223 1,276 1 120 100 40.0 223 1,499
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TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 1 o f 9
The following examples illustrate the design methods presented in the article “Timesaving Design Aids for Reinforced Concrete, Part 3: Columns and Walls,” by David A. Fanella, which appeared in the November 2001 edition of Structural Engineer magazine. Unless otherwise noted, all referenced table, figure, and equation numbers are from that article. The examples presented here are for columns. Examples for walls are available on our Web page: www.portcement.org/buildings. Example 1 In this example, an interior column at the 1st floor level of a 7-story building is designed for the effects of gravity loads. Structural walls resist lateral loads, and the frame is nonsway.
Materials • Concrete: normal weight (150 pcf), ¾-in.
Loads • Floor framing dead load = 80 psf • Superimposed dead loads = 30 psf • Live load = 100 psf (floor), 20 psf (roof) Building Data • Typical interior bay = 30 ft x 30 ft • Story height = 12 ft-0 in. The table below contains a summary of the axial loads due to gravity. The total factored load Pu is computed in accordance with Sect. 9.2.1, and includes an estimate for the weight of the column. Live load reduction is determined from ASCE 7-98. Moments due to gravity loads are negligible.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 2 o f 9
Use Fig. 1 to determine a preliminary size for the tied column at the 1st floor level. Assuming a reinforcement ratio ρg = 0.020, obtain Pu /Ag ≈ 3.0 ksi (f′c = 5 ksi). Since Pu = 1,499 kips, the required Ag = 1,499/3.0 = 499.7 in.2 Try a 22 x 22 in. column (Ag = 484 in.2) with a reinforcement ratio ρg greater than 0.020. Check if slenderness effects need to be considered. Since the column is part of a nonsway frame, slenderness effects can be neglected when the unsupported column length is less than or equal to 12h, where h is the column dimension (Sect. 10.12.2). 12h = 12 x 22 = 264 in. = 22 ft > 12 ft story height, which is greater than the unsupported length of the column. Therefore, slenderness effects can be neglected. Use Fig. 1 to determine the required area of longitudinal reinforcement.
For a 22 x 22 in. column at the 1st floor level: Pu /Ag = 1,499/484 = 3.10 ksi From Fig. 1, required ρg = 0.026, or As = 0.026 x 22 x 22 = 12.58 in.2 Try 8-No. 11 bars (As = 12.48 in.2) Check Eq. (10-2) of ACI 318-99: φPn(max) = 0.80φ[0.85f’c (Ag – Ast) + fy Ast] φPn(max) = 1,542 kips > 1,499 kips O.K. From Table 1, 5-No. 11 bars can be accommodated on the face of a 22-in. wide column with normal lap splices and No. 4 ties. In this case, only 3-No. 11 bars are provided per face. Use 8-No. 11 bars (ρ = 2.58%). Determine required ties and spacing. According to Sect. 7.10.5.1, No. 4 ties are required when No. 11 longitudinal bars are used.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 3 o f 9
According to Sect. 7.10.5.2, spacing of ties shall not exceed the least of: 16 long. bar diameters = 16 x 1.41 16 long. bar diameters = 22.6 in. 48 tie bar diameters = 48 x 0.5 48 tie bar diameters = 24 in. Least column dimension = 22 in. (governs) Check clear spacing of longitudinal bars:
in. 885.6
41.12
241.15.05.1222
spaceClear
=
−
++−
=
Since the clear space between longitudinal bars > 6 in., cross-ties are required per Sect. 7.10.5.3. Reinforcement details are shown below. See Sect. 7.8 for additional special reinforcement details for columns.
22″
22″
8-No. 11
No. 4 ties @ 22″
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 4 o f 9
Example 2 In this example, a simplified interaction diagram is constructed for an 18″ x 18″ tied column reinforced with 8-No. 9 Grade 60 bars (ρg = 8/182 = 0.0247). Concrete compressive strength = 4 ksi. Use Fig. 3 to determine the 5 points on the interaction diagram. • Point 1: Pure compression
kips 871))]485.0(60( 0247.0
)485.0[(1856.0
)]f85.0f(
f85.0[A80.0P
2cyg
cg(max)n
=×−+
××=
′−ρ+
′φ=φ
• Point 2 (fs1 = 0)
Layer 1:
0)1( 11dd
C11
12 =−=−
Layer 2:
42.056.1500.9 11
dd
C11
22 =
−=−
Layer 3:
84.056.1544.2 11
dd
C11
32 =
−=−
Since 1 – C2 (d3 /d1) > 0.69, the steel in layer 3 has yielded. Therefore, set 1 – C2 (d3 /d1) = 0.69 to ensure that the stress in the bars in layer 3 is equal to 60 ksi.
d 3 =
2.4
4″
18″
1.5″
(typ
.)
d 2 =
9.0
0″
d 1 =
15.
56″
18″
No. 3 tie
3-No. 9
2-No. 9
3-No. 9
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 5 o f 9
kips 744)2.2534.809( 70.0
)]}69.03( )42.0(20)7[(38 )1856.1589.2{(70.0
dd
C1A87bdCPn
1i 1
i2si11n
=+=
×+×+×+××=
−+φ=φ ∑
=
{
kips- ft18212/)4.181,11.932,1( 70.0
12/)]}44.29)(69.03( )99)(42.0(2
15.56)0)(97[(38 00.1
56.1585.018
)1856.1589.25.0[(70.0
12/d2h
dd
C1A87
Cd
hbdC5.0M
in
1i 1
i2si
2
1111n
=+=
−×+−×+−×+
×
−×
×××=
−
−+
β−φ=φ
∑=
• Point 3 (fs1 = -0.5fy)
Layer 1:
34.0)1( 34.11dd
C11
12 −=−=−
Layer 2:
23.056.1500.9 34.11
dd
C11
22 =
−=−
Layer 3:
79.056.1544.2 34.11
dd
C11
32 =
−=−
Use 0.69
kips 514)4.1312.602(70.0
)]}69.03( )23.0(2)34.0-7[(38
)1856.1515.2{(70.0
dd
C1A87bdCPn
1i 1
i2si11n
=+=
×+×+×+
××=
−+φ=φ ∑
=
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 6 o f 9
{
kips- ft24612/)5.763,18.447,2( 70.0
12/)]}44.29)(69.03( )99)(23.0(2
15.56))(934.0-7[(38 34.1
56.1585.018
)1856.1515.25.0[(70.0
12/d2h
dd
C1A87
Cd
hbdC5.0M
in
1i 1
i2si
2
1111n
=+=
−×+−×+
−×+
×
−×
×××=
−
−+
β−φ=φ
∑=
• Point 4 (fs1 = -fy)
Layer 1:
69.0)1( 69.11dd
C11
12 −=−=−
Layer 2:
02.056.1500.9
69.11dd
C11
22 =
−=−
Layer 3:
74.056.1544.2
69.11dd
C11
32 =
−=−
Use 0.69
kips 338)5.39.478( 70.0)]}69.03(
)02.0(2)69.0-7[(38 )1856.1571.1{(70.0
dd
C1A87bdCPn
1i 1
i2si11n
=+=×+
×+×+××=
−+φ=φ ∑
=
{
kips- ft28012/)8.362,23.436,2( 70.0
12/)]}44.29)(69.03( )99)(02.0(2
15.56))(969.0-7[(38 69.1
56.1585.018
)1856.1571.15.0[(70.0
12/d2h
dd
C1A87
Cd
hbdC5.0M
in
1i 1
i2si
2
1111n
=+=
−×+−×+
−×+
×
−×
×××=
−
−+
β−φ=φ
∑=
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 7 o f 9
• Point 5: Pure bending
Use iterative procedure to determine φMn. Try c = 4.0 in.
0087.04
56.154 003.0
cdc
003.0 11s
−=
−
=
−
=ε
kips 180)60(3fAT
ksi 60 fuse , ksi60- ksi 4.251)0087.0(000,29
Ef
1s1ss1
s1
1ss1s
−=−×==
−=>−=−×=
ε=
0038.04
94 003.0
cdc
003.0 12s
−=
−
=
−
=ε
kips 120)60(2fAT
ksi 60 fuse , ksi60-ksi 8.108)0038.0(000,29
Ef
2s2ss2
s2
2ss2s
−=−×==
−=>−=−×=
ε=
0012.04
44.24 003.0
cdc
003.0 13s
=
−
=
−
=ε
kips 1029.333fAC
ksi 9.330012.0000,29Ef
3s3ss3
3ss3s
=×==
=×=ε=
kips 20818)485.0(485.0
abf85.0C cc
=××××=
′=
Total T = (-180) + (-120) = -300 kips Total C = 102 + 208 = 310 kips Since T ≈ C, use c = 4.0 in.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 8 o f 9
kips- ft4.98
12/56.15218)180(
d2h
TM 11s1ns
=
−−=
−=
0
12/9218
)120(
d2h
TM 22s2ns
=
−−=
−=
kips- ft8.55
12/44.2218 102
d2h
CM 33s3ns
=
−=
−=
kips- ft7.2802.15412/)]4.318(2085.0[
M)ah(C5.0M3
1insicn
=+−××=
+−= ∑=
kips- ft2537.2809.0Mn =×=φ
Compare simplified interaction diagram to interaction diagram generated from the PCA computer program PCACOL. The comparison is shown on the next page. As can be seen from the figure, the comparison between the exact (black line) and simplified (red line) interaction diagrams is very good.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS Columns
P o r t l a n d C e m e n t A s s o c i a t i o n P a g e 9 o f 9
